Help Solve My Non-Calculus Physics Homework Problem

[liliphys]

I have a HW problem for non-calculus based physics that I can't seem to figure out. Please help me if you can:

Consider a sphere of radius R woth a uniform positive charge distribution (rho)=q/V. Using Gauss's law, prove (a) that the magnitude of the electric field outside the spherical volume is E=q/4(pi)(permittivity of free space)r^2= (rho) R^3 / 3 (permittivity of free space)r^2, (b) that the magnitude of the electric field insode the sphere at a distance r from the center is E=(rho)r/(permittivity), (c) compare the results of a and b when r=R.


I have Gauss's law: E(4(pi)r^2=4(pi)koQ and that for r>R, E=ko Q/r^2
and i substituted in (rho)V for Q and 4/3 (pi)r^3 for V and simplified, but this did not prove part A.
And for r<R, Q'=Q(r/R)^3 because the inner sphere encloses a volume 4/3 (pi)r^3 which is (r/R)^3 times the total volume. but this doesn't work out either. And how do i reconcile ko from coulomb with the permittivity?

 

[Doc Al]

Originally posted by liliphys
I have Gauss's law: E(4(pi)r^2=4(pi)koQ and that for r>R, E=ko Q/r^2
and i substituted in (rho)V for Q and 4/3 (pi)r^3 for V and simplified, but this did not prove part A.

All you need is Gauss's law. Gauss's law in terms of permitivity of free space (pfs) is:
E*(Area) = Q/(pfs). The area is 4&pi;r²; the volume of charge is 4/3*&pi;R³. This should give you the answer you seek.

And for r<R, Q'=Q(r/R)^3 because the inner sphere encloses a volume 4/3 (pi)r^3 which is (r/R)^3 times the total volume. but this doesn't work out either.

For r<R, use the same idea. The area is 4&pi;r²; the volume of charge is 4/3*&pi;r³. This should work.

And how do i reconcile ko from coulomb with the permittivity?

You don't need to use Coulomb's law, only Gauss's law. But k0 = 1/(4&pi;(pfs)).

 

[M98Ranger]

First of all, don't worry, I am here to help you solve this problem. Let's start with part A: To prove that the magnitude of the electric field outside the spherical volume is E=q/4(pi)(permittivity of free space)r^2= (rho) R^3 / 3 (permittivity of free space)r^2, we can use Gauss's law. According to Gauss's law, the electric flux through any closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space. In this case, we have a spherical surface with radius r, outside the spherical volume. Therefore, the electric flux through this surface is given by E(4(pi)r^2)=q/permittivity of free space. Rearranging this equation, we get E=q/4(pi)(permittivity of free space)r^2, which is the desired result.

Moving on to part B: To prove that the magnitude of the electric field inside the sphere at a distance r from the center is E=(rho)r/(permittivity), we can again use Gauss's law. In this case, we have a spherical surface with radius r, inside the spherical volume. Therefore, the electric flux through this surface is given by E(4(pi)r^2)=q'/permittivity of free space, where q' is the charge enclosed by this surface. Now, q' is equal to the charge density (rho) multiplied by the volume enclosed by the surface, which is 4/3 (pi)r^3. Therefore, q'=(rho)(4/3)(pi)r^3. Substituting this value in the equation, we get E(4(pi)r^2)=((rho)(4/3)(pi)r^3)/permittivity of free space. Simplifying this equation, we get E=(rho)r/permittivity of free space, which is the desired result.

Lastly, let's compare the results of part A and B when r=R. In part A, we get E=q/4(pi)(permittivity of free space)R^2. In part B, we get E=(rho)R/permittivity of free space. Now, we know that q=(rho)V, where V is the volume of the sphere. In this case, V=4/3 (pi)R^3. Substituting